Messages posted by vnkt.swaroop

soulubility increases either with increase in pressure or with decrease with temperature.

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dipole moment is the product of charges and the distance between the atoms of the molecule.

polarity is just the charge that developed on the atoms in a molecule due to diference in electronegativities of the atoms

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(x)log(z)=(2x)log(y) (z)log(2)=log(2)+(2x)log(2)

x=0 or z=y^2 z=1+2x

if x=0 then z=1 and y=1 or y=(-1) this is not possible as it is given x+y+z=16.

and hence 2x+1=y^2=z

x+y+z=16

2x+2y+2z=32

(2x+1)+2y+2z=33

y^2+2y+2y^2=33

2y^(2)+2y-33=0

y={-1+(root67)}/{2} or y={-1-(root67)}/{2}

and now get "x" and "z".

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if we know its magnetic moment then we can say whether the Copper atom is having unpaired electron or not.

if you any more doubts please post me on swaroop.exlnc@gmail.com and i will clarify your doubts.

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1/(sinx-4)(sinx-1)

substitute Sinx={2tan(x/2)}/{1+tan^2(x/2)}

now take tan(x/2)=t and solve.

if you not getting the sum tell me then i will post the solution.

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{SinA}{[1-Sin^2(A)]^(1/2)}=1/2

{Sin^(2)A}{[1-Sin^(2)A]}=1/4 (squaring on both sides)

t(1-t)=1/4 (let Sin^(2)A=t)

t^(2)-t+(1/4)=0

{t-(1/2)}^(2)=0

t=1/2

Sin^(2)A=1/2

SinA={1/2}^(1/2) or SinA=-{1/2}^(1/2)

therefore SinA+CosA=+(root2) , -(root2)

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if we reduce the compound then it is correct otherwise both are having zero stereocentric centers.

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yes.of course.

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a carbon atom which is saturated can only be considered as stereogenic center, and a "carbon" atom which is having double or triple bonds cannot be considered as a stereogenic center.

Stereogenic Center can also be reffered as a Chiral Center (chiral carbon atom). in orer to be a stereogenic center a carbon atom should have all the 4 groups attached to it different from each other (All The Groups Should Be Different).

now you try and find out how many Stereogenic Centers are present in the compound. If you are not getting then repyl to me and i will post you the solution.

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secx/sec2x

=cos2x/cosx

={2cos^2(x)-1}/{cosx}

=2cos(x)-secx

=2sin(x)-ln(secx+tanx)

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substitute y=vx then dy/dx=v+x(dv/dx)

v+x(dv/dx)=(x^2-x^2v^2)/(x^2+x^2v^2)

v+x(dv/dx)=(1-v^2)/(1+v^2)

now solve and try to get it.

if you are not able to get it, tell me i will post the post the entire solution.

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who said it is not differentiable.

it is differentiable. F(x) is a continous and defined at x=0.

LHD(Left Hand Derrivative)=-2x

RHD(Right Hand Derrivative)=2x but on substituting the values of x we get LHD=RHD.

Hence F(x) is diffenrentiable at x=0.

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the hybridissation of Oxygen in H2O2 is sp3 (2lone pairs+ 2 bond pairs).

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the type of bonds are ionic and covalent.

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y=e^(x) x=0 y=e constitute the area present under y=e^(x) and y-axis.

y=e implies x=1

this area is = (rectangle formed by y=e and x=1) - (area under y=e^(x) under the limits 0 and 1)

therefore area = (e) - (e-1)

=1squareunit

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{(cosx+cos7x)+(cos2x+cos6x)+(cos3x+cos5x)+(cos4x)}/{(sinx+sin7x)+(sin2x+sin6x)+(sin3x+sin5x)+(sin4x)}

={(2cos4xcos3x)+(2cos4xcos2x)+(2cos4xcosx)+(cos4x)}/{(2sin4xcos3x)+(2sin4xcos2x)+(2sin4xcosx)+(sin4x)}

=(cos4x){2cos3x+2cos2x+2cosx+1}/(sin4x){2cos3x+2cos2x+2cosx+1}

now check and find the answer. if you are not getting tell me i will tell the final solution.

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e^sinx=t

t-(1/t)-4=0

t^2-4t-1=0

t={4+2root(5)}/{2} t={4-2root(5)}/{2}

t=2+root(5) t=2-root(5)

now check and see whether it is correct ot not?

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the product is m-nitro nitrobenzene.

as nitro group is meta directing the incoming NO2(+) group comes and attacks at meta position and as the already present nitro group is electron withdrawing group the reaction is almost not possible.

or else we can say that "The Reaction Is PossibleWith Very Little Amount Of Product (nearly 0.1%)".

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Pb3O4 exists as PbO and PbO2

2PbO + PbO2 = Pb3O4 (it is called Red Lead or Sindhur )

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